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  1. Mar 26, 2005 #1
    if G is a group of order 2n then show that it has an element of order 2 ( and odd number of them)
    i've been thinking about this...and i think i've gotten somewhere...
    e belongs to G and o(e) = 1
    now if there's no element of order 2 in G...we're looking at elements which are not their own inverses...such elements come in pairs, so i guess there lies the contradiction...
    i don't have a clue as to how to show that there are odd number of elements of order two.
  2. jcsd
  3. Mar 26, 2005 #2

    matt grime

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    just carry on in the same vein.

    G as a set is the union of the pairs {x,x^{-1}} where these are distinct, with the set of elements of order 2 and the set containing the identity. equating orders

    |G| =2n = 2K + L + 1

    where K is the number if pairs, and L is the number of self inverse non-identity elements.
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